# How do you find the roots, real and imaginary, of y= -3x^2 -2x +(2x+1)^2  using the quadratic formula?

Jul 24, 2018

$y = - 3 {x}^{2} - 2 x + {\left(2 x + 1\right)}^{2}$ has a double root at $x = - 1$

#### Explanation:

$y = - 3 {x}^{2} - 2 x + {\left(2 x + 1\right)}^{2}$
$y = - 3 {x}^{2} - 2 x + \left(4 {x}^{2} + 4 x + 1\right)$
$y = - 3 {x}^{2} - 2 x + 4 {x}^{2} + 4 x + 1$
$y = {x}^{2} + 2 x + 1$

Using the quadratic formula
$x = \frac{- 2 \pm \sqrt{4 - 4 \left(1\right) \left(1\right)}}{2}$
$x = \frac{- 2 \pm \sqrt{0}}{2}$
$x = \frac{- 2}{2}$
$x = - 1$

Therefore, $y = - 3 {x}^{2} - 2 x + {\left(2 x + 1\right)}^{2}$ has a double root at $x = - 1$